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Aluminum metal reacts rapidly with aqueous sulfuric acid to produce aqueous aluminum sulfate and hydrogen gas: 2 Al (s) + 3H,SO, (ag) -> Al (SOg) (ag) + 3H, (8) Determine the volume of hydrogen gas produced at STP when a 2.00 g piece of aluminum completely reacts.

User Docwhat
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Final answer:

The volume of hydrogen gas produced when 2.00 g of aluminum reacts with sulfuric acid at STP is calculated to be 2.4901 liters, using stoichiometry and the molar volume of a gas at STP.

Step-by-step explanation:

The volume of hydrogen gas produced when a 2.00 g piece of aluminum reacts completely with aqueous sulfuric acid to form aqueous aluminum sulfate can be determined through stoichiometric calculations. At STP (standard temperature and pressure), one mole of a gas occupies 22.4 liters. First, we convert the mass of aluminum to moles using its molar mass (atomic weight of Al is approximately 26.98 g/mol):

2.00 g Al * (1 mol Al / 26.98 g Al) = 0.0741 moles Al

According to the balanced chemical equation 2 Al (s) + 3 H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + 3 H₂ (g), 2 moles of Al produce 3 moles of H₂ gas.

So, 0.0741 moles Al * (3 moles H₂ / 2 moles Al) = 0.1112 moles H₂

Finally, convert moles of hydrogen to volume at STP:

0.1112 moles H₂ * (22.4 L / 1 mol H₂) = 2.4901 liters

Therefore, 2.4901 liters of hydrogen gas are produced when 2.00 g of aluminum reacts with sulfuric acid at STP.

User Aleksei Kornushkin
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